Lemma 5

(i) $\mathscr{U}_+$ is a lattice in the vector space $U(\mathfrak{n}_+)$.

(ii) For all $\alpha \in B$, we have $\mathscr{U}_+\cap U(\mathfrak{g}^{\alpha}) =\bigoplus_{n\in\mathbf{N}}\mathbf{Z}x^{(n)}_{\alpha}$.

By definition, $\mathscr{U}_+$ is generated as a $\mathbf{Z}$-module by the elements

$x(\varphi \mathbf{n})$ = $1\leq \prod i\leq rx(\varphi n((ii)$))

where $r\in \mathbf{N},\varphi = (\varphi (i))\in B^r$, and $\mathbf{n}= (n(i))\in \mathbf{N}^r$. Give the algebra $U(\mathfrak{n}_+)$ the graduation of type Q(R) for which each $\mathfrak{g}^{\alpha}(\alpha \in R_+)$ is homogeneous of degree $\alpha$. A monomial $x^{(\mathbf{n})}_{\varphi}$ of the preceding type is homogeneous of degree

The monomials of this kind having a given degree $q$ are finite in number, and generate over $\mathbf{Q}$ the homogeneous component of $U(\mathfrak{n}_+)$ of degree $q$. This proves (i).

If $\alpha \in B,\mathscr{U}_+\cap U(\mathfrak{g}^{\alpha})$ is contained in the sum of the homogeneous components of degrees which are multiples of $\alpha$; thus, by the preceding, $\mathscr{U}_+\cap U(\mathfrak{g}^{\alpha})$ is generated by the $x^{(\mathbf{n})}_{\varphi}$ such that $\sum n(i)\varphi (i)\in \mathbf{N}\alpha$, which forces $\varphi (i) =\alpha$ for all $i$ (since B is a basis of R), so

$x^{(\mathbf{n})}_{\varphi}=x^{(n(1))}_{\alpha}. . . x^{(n(r))}_{\alpha}=(n(1) +n$(1)!$\cdots . . . n+(nr$)!$(r$))! $x^{(n(1)+\cdot \cdot \cdot+n(r))}_{\alpha}$.

Thus, $\mathscr{U}_+\cap \mathscr{U}(\mathfrak{g}^{\alpha})\subset \bigoplus_n\mathbf{Z}x^{(n)}_{\alpha}$, hence (ii). Q.E.D.

In the remainder of this paragraph, if E and F are $\mathbf{Z}$-submodules of $U(\mathfrak{g})$, we denote by $E.F$ the $\mathbf{Z}$-submodule of $U(\mathfrak{g})$ generated by the products $ab$, where $a\in E, b\in F$.